Two identical parallel plate capacitors,of ca | vedclass

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(B) For capacitor $(I)$,the dielectrics are in series. The capacitance of each part is $C_i = \frac{3 \varepsilon_0 A K_i}{d}$.Since they are in serie

Vedclass · Accepted Answer

$\frac{E_1}{E_2} = \frac{9 K_1 K_2 K_3}{(K_1 + K_2 + K_3)(K_2 K_3 + K_3 K_1 + K_1 K_2)}$

Vedclass · Answer

(B) For capacitor $(I)$,the dielectrics are in series. The capacitance of each part is $C_i = \frac{3 \varepsilon_0 A K_i}{d}$.Since they are in series,$\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} = \frac{d}{3 \varepsilon_0 A} (\frac{1}{K_1} + \frac{1}{K_2} + \frac{1}{K_3}) = \frac{d(K_2 K_3 + K_3 K_1 + K_1 K_2)}{3 \varepsilon_0 A K_1 K_2 K_3}$.Thus,$C_{eq} = \frac{3 \varepsilon_0 A K_1 K_2 K_3}{d(K_1 K_2 + K_2 K_3 + K_3 K_1)}$.For capacitor $(II)$,the dielectrics are in parallel. The capacitance of each part is $C_i' = \frac{\varepsilon_0 (A/3) K_i}{d} = \frac{\varepsilon_0 A K_i}{3d}$.Since they are in parallel,$C_{eq}' = C_1' + C_2' + C_3' = \frac{\varepsilon_0 A}{3d} (K_1 + K_2 + K_3)$.The energy stored is $E = \frac{1}{2} C V^2$. Therefore,$\frac{E_1}{E_2} = \frac{C_{eq}}{C_{eq}'} = \frac{3 \varepsilon_0 A K_1 K_2 K_3}{d(K_1 K_2 + K_2 K_3 + K_3 K_1)} \cdot \frac{3d}{\varepsilon_0 A (K_1 + K_2 + K_3)} = \frac{9 K_1 K_2 K_3}{(K_1 + K_2 + K_3)(K_1 K_2 + K_2 K_3 + K_3 K_1)}$.

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